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Since we were given that the roots were positive reals,we get motivated and tempted to use,\(R.M.S\geq A.M\geq G.M\geq H.M\),also since we have been given,\(\sum \alpha\) and \(\sum \alpha\beta\) we try to find the value of \(\sum \alpha^2=4^2-2\times 7=2\).We apply \(R.M.S \geq A.M\)
\[\sqrt{\dfrac{2}{8}} \geq \dfrac{4}{8}\\
\Longrightarrow \dfrac{1}{2} \geq \dfrac{1}{2}\],hence equality occurs,meaning that all the roots are equal.Now,that means,\(A.M=G.M\),hence\[\dfrac{4}{8}=\sqrt[8]{f}\\
\Longrightarrow \dfrac{1}{256}=f\],hence there is only one value of \(f=\dfrac{1}{256}\).Is this correct Harsh?
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Adarsh Kumar
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1 year ago

@Anish Harsha
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U may use the concept of fermat pt to solve this one..then it will be very easy...suppose three line segments AF,BF,CF of length x,y,z meet each other at120 degrees and now use cosine rule to find out the lengths of the triangle and see that each term on the rhs of the inequality is actually squares of the sides...then express xy+yz+zx as \(\frac{4}{\sqrt{3}}T\) where T is the area of the triangle....then use \(T=\frac{abc}{4R}\) and see that the inequality reduces to 3R greater than or equal to (x+y+z) which is obvious since F is the fermat pt
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Souryajit Roy
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1 year ago

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You have a set of consecutive positive integers,\(\{1,2,3,...,199,200\}\),if you choose \(101\) numbers from these at random,then prove that there at-least two numbers of which one divides the other.
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Adarsh Kumar
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1 year ago

@Shrihari B
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Yes , Menelaus is indeed a good approach for this problem. For shortening the solution: Once you get F is the midpoint of CD , you can directly say that \(\angle BAC = 90^\circ\) , do you see why?
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Nihar Mahajan
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1 year ago

@Harsh Shrivastava
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Take LCM on the LHS and try proving that each term in the numerator leaves a different residue modulo p. Do it by contradiction. Once you are able to prove that, the rest is clear. If it is difficult, first take a smaller example say p=7. Then work it out
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Shrihari B
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1 year ago

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@Shrihari B
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I have the solution only till \(a\) is divisible by \(p\). To prove that \(a\) is divisible by \(p^{2}\) is given below the solution as a challenge.
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Svatejas Shivakumar
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1 year ago

We note that the set \(S\) can be partitioned into two sets \(S_1\) and \(S_2\) according as \(qx>py\) or \(qx<py\). Note that there are no pairs in such \(S\) such that \(qx=py\).

The set \(S_1\) can be described as the set of all pairs \((x,y)\) satisfying \(1 \le x \le (p-1)/2, \quad 1 \le y < qx/p\). Then \(S_1\) has \(\displaystyle \sum_{j=1}^{(p-1)/2}{\frac{qj}{p}}\) elements.

From here, it is clear that either \(x=2\) or \(y=2\). We now separate this into 2 cases:

Case 1: \(y=2\)

When \(y=2\), we have \(19 - 2 = 17 \equiv 0 \pmod{x}\), which implies \(x=17\). The solution in this case is \((17, 2)\). Checking, however, we find that this clearly doesn't work. So, there are no solutions in this case.

Case 2: \(x=2\)

When \(x=2\), we have \(19+2 = 21 \equiv 0 \pmod {y}\), which implies \(y=3\) or \(y=7\). Checking, we find both these solutions work. Thus, 2 solutions exist in this case: \((2, 3)\) and \((2,7)\).

If \(a,b,c\) are positive integers such that \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}<1\). Prove that the maximum value of \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) is \(\dfrac{41}{42}\)
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Svatejas Shivakumar
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1 year ago

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@Svatejas Shivakumar
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If none of \(a,b,c=2\), the maximum value of the expression will be \(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}<\frac{41}{42}\). So, one of them has to \(2\) say \(a\). So we have to prove that the maximum value of \(\frac{1}{b}+\frac{1}{c}\) is \(\frac{10}{21}\). If none of \(b,c=3\), the maximum of the expression will be \(\frac{1}{4}+\frac{1}{5}=\frac{9}{20}<\frac{10}{21}\). So, one them has to be \(3\) say \(b\). Therefore, \(\frac{1}{c} \le \frac{1}{7}\).Therefore the maximum value of \(c\) is \(7\) and the maximum value of \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) is \(\frac{41}{42}\).

@Harsh Shrivastava
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Since the expression is symmetric, let's assume without loss of generality that \(x \ge y\).

\(\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge \frac{y^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 4+\frac{y^{2}}{x-1} \implies \frac{x^{2}}{y-1} \ge 4\). Minimum value of \(\frac{x^{2}}{y-1}\) is \(4\) and it occurs if and only if \(x=y=2\).

Everyone here who have posted questions on this board and have not got any solutions please post the solutions now as RMO is only 42 hours from now. Its better that we understand the methods.
Sorry that I cannot send the solution of my geometry question as I myself have been trying that question but could not arrive at the solution.

And please post some more questions if possible :)
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Shrihari B
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1 year ago

@Saarthak Marathe
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Hey Saarthak I haven't heard of Catalan numbers before. You need to elaborate a bit
Could you give a proof for the number of ways ? It will help me understand better. Thanks in advance
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Shrihari B
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1 year ago

Here is a geometry question. Waiting for numerous solutions.
In a triangle ABC, excircle opposite to A touches BC at point Q. Let M be the midpoint of BC. Incircle of triangle ABC touches BC at point P. Prove that PM=MQ.
I have been trying this question but I am stuck. So please solve this question.
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Shrihari B
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1 year ago

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@Shrihari B
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Its easy , note that \(BP=QC=s-b\) where \(s\) is the semiperimeter of the triangle and \(b\) is length of side opposite to \( \angle B\) respectively. Since \(M\) is mid point of \(BC\) , the result follows.
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Nihar Mahajan
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1 year ago

Let d(k) denote the sum of the digits of k in base10. Find all natural numbers n such that n, d(n), d(d(n)) sum to 2015.
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Rakhi Bhattacharyya
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1 year ago

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@Rakhi Bhattacharyya
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I am getting no solution. Is that right ?
My proof : n,d(n),d(d(n)) are congruent to the same number say 'x' (mod 9). So their sum is congruent to 3x (mod 9). Therefore 2015 is congruent to 3x ( mod 9) which is impossible as 3 does not divide 2015
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Shrihari B
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1 year ago

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Find all pairs \((m,n)\) such that \(2^m+3^n\) is a perfect square. (A popular NT problem)
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Nihar Mahajan
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1 year ago

@Nihar Mahajan
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Taking mod 4 we get that m,n should be even. Then using \( gcd({2}^{m},3^{n})=1 \), we use the general formula for pythagorean triplets.
Taking various cases,we get that only \( m=4,n=2\) are the solutions.
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Saarthak Marathe
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1 year ago

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If \(m\) is a positive real number and satisfies the equation \[2m=1+\frac{1}{m}+\sqrt{m^{2}-\frac{1}{4}}\] Find \(m\).
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Shivam Jadhav
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1 year ago

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@Shivam Jadhav
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Could you send the solution of this question. I tried expanding and landed up with a fourth degree polynomial whose roots I am unable to guess
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Shrihari B
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1 year ago

@Saarthak Marathe
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First use an image uploading website (I prefer postimg.org ), upload your picture, and copy the image link. The format for uploading images in a solution is

Alternate text

. Type the text which you wish to be displayed if the image doesn't load in the third brackets, and paste the image URL in the second brackets. In the following example, the image URL is http://s12.postimg.org/5qhlgca71/untitled.png, and I wrote the code

@Svatejas Shivakumar
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Well the question becomes very easy if u make it a quadratic in b and then prove that the Discriminant is less than 0. I did it that way.
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Shrihari B
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1 year ago

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An easy INMO geometry problem:

Let \(ABC\) be a triangle with circumcircle \(\Gamma\). Let \(M\) be a point in the interior of triangle \( ABC\) which is also on the bisector of \(\angle A\). Let \(AM, BM, CM\) meet \( \Gamma\) in \( A_{1}, B_{1}, C_{1}\) respectively. Suppose \(P\) is the point of intersection of \( A_{1}C_{1}\) with \(AB\) and \(Q\) is the point of intersection of \(A_{1}B_{1}\) with \(AC\). Prove that \(PQ\) is parallel to \(BC\) .
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Nihar Mahajan
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1 year ago

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Let \(ABC\) be an acute-angled tiangle .Let O denote its circumcenter.Let \(\Pi\) be the circle passing through the points A,O,B.Lines CA and CB meet \(\Pi\) again at P and Q respectively.Prove that PQ is perpendicular to the line CO.
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Harsh Shrivastava
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1 year ago

Let \(ABC\) be a triangle and \(P\) be a point inside it.Let \(x,y,z\) denote the lengths of \(PA\),\(PB\),\(PC\).Let \(a,b,c\) be the lengths of sides \(BC\),\(CA\),\(AB\).Find all points \(P\) such that \(axy+byz+czx=abc\).
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Souryajit Roy
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1 year ago

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If a circle goes through point \(A\) of a parallelogram \(ABCD\), cuts the two sides \(AB,AD\) and the diagonal \(AC\) at points \(P,R\) and \(Q\) respectively. Prove that \((AP×AB)+(AR×AD)=AQ×AC\).
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Svatejas Shivakumar
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1 year ago

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@Svatejas Shivakumar
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I hope you all have drawn the diagram.(Consider all the segments below as vectors in the given direction.)

Let \(AS\) be the diameter of given circle.

\(AB+AD=AC\)

Therefore taking dot product with\(AS\)

\(AB.AS+AD.AS=AC.AS \)

Using, \(a.b=|a|*|b|*cos(\theta) \) [\( \theta\) is angle between the vectors]

@Saarthak Marathe
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You can also use Ptolemy's theorem to the quadrilateral \(PQRA\) and then replace the sides of \(\Delta PQR\) by the corresponding sides of the similar triangle \(CBA\).( \(\Delta PQR \sim \Delta CBA\) by \(AAA\) similarity.)
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Svatejas Shivakumar
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1 year ago

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Given seven arbitrary distinct real numbers, show that there exist two numbers \(x\) and \(y\) such thst \(0<\dfrac{x-y}{1+xy}<\dfrac{1}{\sqrt{3}}\).
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Svatejas Shivakumar
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1 year ago

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@Svatejas Shivakumar
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The question is easy. Consider x and y to be some tan(A) and tan(B) respectively and consider A and B to vary from 0 to pi so that all reals are covered. now divide it into six equal portions spaced by pi/6. Now u have 6 pigeons and 7 holes and you are done. Then that expression is simply tan(A-B).
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Shrihari B
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1 year ago

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There are given \(2^{500}\) points on a circle labeled \(1, 2, . . . , 2^{500}\)in some order. Prove that
one can choose 100 pairwise disjoint chords joining some of these points so that the 100 sums
of the pairs of numbers at the endpoints of the chosen chords are equal.
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Lakshya Sinha
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1 year ago

@Svatejas Shivakumar
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Hey !! That doesn't stop you from taking part in the discussions on INMO. Its okay ... Ur just in 8th standard, I came to know about RMO in 10th ...so u are at an advantage. And I am quite sure u will make it to the IMO by your 11-12th standard. I just don't know how u r able to solve olympiad questions being in 8th standard. So now bro .... u r targeting for IMO and thus u r participating in the INMO board ...
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Shrihari B
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11 months, 2 weeks ago

@Shrihari B
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Congratulations!! The results have not been declared from my region (I don't think that I have much chances of qualifying this time. I did many silly mistakes :(
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Svatejas Shivakumar
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12 months ago

Sorry didn't see this post but since I found it, here you go - 65 bugs are placed at different squares of a 9*9 square board. Abug in each moves to a horizontal or vertical adjacent square. No bug makes 2 horizontal or vertical moves in succession. Show that after some moves, there will be at least 2 bugs in the same square.
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Satyajit Ghosh
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1 year ago